3.7.23 \(\int x^3 (a+b x^4)^2 \, dx\) [623]

Optimal. Leaf size=16 \[ \frac {\left (a+b x^4\right )^3}{12 b} \]

[Out]

1/12*(b*x^4+a)^3/b

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Rubi [A]
time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \begin {gather*} \frac {\left (a+b x^4\right )^3}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^4)^2,x]

[Out]

(a + b*x^4)^3/(12*b)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x^3 \left (a+b x^4\right )^2 \, dx &=\frac {\left (a+b x^4\right )^3}{12 b}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 30, normalized size = 1.88 \begin {gather*} \frac {a^2 x^4}{4}+\frac {1}{4} a b x^8+\frac {b^2 x^{12}}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^4)^2,x]

[Out]

(a^2*x^4)/4 + (a*b*x^8)/4 + (b^2*x^12)/12

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Maple [A]
time = 0.14, size = 15, normalized size = 0.94

method	result	size
default	\(\frac {\left (b \,x^{4}+a \right )^{3}}{12 b}\)	\(15\)
gosper	\(\frac {1}{12} b^{2} x^{12}+\frac {1}{4} a b \,x^{8}+\frac {1}{4} a^{2} x^{4}\)	\(25\)
norman	\(\frac {1}{12} b^{2} x^{12}+\frac {1}{4} a b \,x^{8}+\frac {1}{4} a^{2} x^{4}\)	\(25\)
risch	\(\frac {b^{2} x^{12}}{12}+\frac {a b \,x^{8}}{4}+\frac {a^{2} x^{4}}{4}+\frac {a^{3}}{12 b}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^4+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/12*(b*x^4+a)^3/b

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Maxima [A]
time = 0.29, size = 14, normalized size = 0.88 \begin {gather*} \frac {{\left (b x^{4} + a\right )}^{3}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^4+a)^2,x, algorithm="maxima")

[Out]

1/12*(b*x^4 + a)^3/b

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Fricas [A]
time = 0.36, size = 24, normalized size = 1.50 \begin {gather*} \frac {1}{12} \, b^{2} x^{12} + \frac {1}{4} \, a b x^{8} + \frac {1}{4} \, a^{2} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^4+a)^2,x, algorithm="fricas")

[Out]

1/12*b^2*x^12 + 1/4*a*b*x^8 + 1/4*a^2*x^4

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (10) = 20\).
time = 0.01, size = 24, normalized size = 1.50 \begin {gather*} \frac {a^{2} x^{4}}{4} + \frac {a b x^{8}}{4} + \frac {b^{2} x^{12}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**4+a)**2,x)

[Out]

a**2*x**4/4 + a*b*x**8/4 + b**2*x**12/12

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Giac [A]
time = 0.71, size = 14, normalized size = 0.88 \begin {gather*} \frac {{\left (b x^{4} + a\right )}^{3}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^4+a)^2,x, algorithm="giac")

[Out]

1/12*(b*x^4 + a)^3/b

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Mupad [B]
time = 0.03, size = 24, normalized size = 1.50 \begin {gather*} \frac {a^2\,x^4}{4}+\frac {a\,b\,x^8}{4}+\frac {b^2\,x^{12}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^4)^2,x)

[Out]

(a^2*x^4)/4 + (b^2*x^12)/12 + (a*b*x^8)/4

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